Vol 6, No 10 (2015)

(1)

Available online throug

ISSN 2229 – 5046

SOLUTION OF LINEAR HIGHER, NONLINEAR HIGHER ORDER BOUNDARY VALUE

PROBLEMS WITH FRACTIONAL ORDER FUNCTION BY USING

DIFFERENTIAL TRANSFORM METHOD

AVINASH KHAMBAYAT*

1

_{, NARHARI PATIL}

2

1

_{Department of Applied Science,}

Shram Sadhana Bombay Trusts College of Engineering &Technology,

Bambhori-Jalgaon, Maharashtra, India.

2

_{Professor & Head, Department of Mathematics}

Shri Sant Gajanan Maharaj College of Engineering, Shegaon, Maharashtra, India.

(Received On: 07-10-15; Revised & Accepted On: 29-10-15)

ABSTRACT

I

n this paper we solve some numerical examples of linear and nonlinear higher order with fractional order function by using Differential Transform Method, it gives the convergent series form effectively and the result obtains by Differential Transform Method shows the approximate solution of the Differential equations.

Keywords: Series solutions, Differential transform Method, boundary value problem, Numerical solution.

INTRODUCTION

A variety of methods, exact, approximate and purely numerical are available for the solution of differential equations. Most of these methods are computationally intensive because they are trial-and error in nature, or need complicated symbolic computations. The differential transformation technique is one of the numerical methods for ordinary differential equations. The concept of differential transformation was first proposed by Zhou in 1986 [15] and Arikhoglu[1] and Ozkol, Ayaz [3], Chen [6-7] and Ho, 1996, 1999; Hassan[9] and Abdel-Halim[2]2008, Duan [8], Khaled Batiha[10], it was applied to solve linear and nonlinear initial value problems in electric circuit analysis, Kho, Mantri [11-12], Bert [4].This method constructs a semi analytical, numerical technique that uses the Taylor series for the solution of differential equations in the form of a polynomial. It is different from the high-order Taylor series method which requires symbolic computation of the necessary derivatives of the data functions. The Taylor series method is computationally time-consuming, especially for higher order equations [5].The differential transform is an iterative procedure for obtaining analytic Taylor series solutions of differential equations. The Differential transformation method is very effective and powerful for solving various kinds of differential equation.

In this paper, we solve some numerical examples of linear, nonlinear higher order and fractional order boundary value problem Shahid [13], V. Ananathaswamy [14] by using Differential Transform Method the numerical results are compared with their exact solutions. The main advantage of the DTM is that, it can be applied directly to linear and nonlinear ordinary differential equations without linearization; it reduces the size of computational work and providing the series solution with convergence.

BASIC IDEAS OF DIFFERENTIAL TRANSFORM METHOD

The transformation of the 𝑘𝑘𝑡𝑡ℎ Derivative of a function with one variable follows:

0

) ( ! 1 ) (

x x k k

dx x u d k k U

=

   

 

= (1)

Where 𝑢𝑢(𝑥𝑥) is the original function and 𝑈𝑈(𝑘𝑘)is the transformed function and the differential inverse transformation

𝑈𝑈(𝑘𝑘) is defined by,

∑

∞

=

− =

0

0)

)( ( )

( k

k x x k U x

u (2)

(2)

When 𝑥𝑥0= 0 , the function 𝑢𝑢(𝑥𝑥) Defined in (2) is expressed as

∑

=∞ − =k k k x k U x u 0 ) ( )

( (3)

Equation (3) implies that the concept of one dimensional differential transform is almost is same as the one dimensional Taylors series expansion. We use following fundamental theorems on differential transform method

Theorem 1: If 𝑢𝑢(𝑥𝑥) = 𝛼𝛼𝛼𝛼(𝑥𝑥) ±𝛽𝛽ℎ(𝑥𝑥) 𝑡𝑡ℎ𝑒𝑒𝑒𝑒 𝑈𝑈(𝑘𝑘) =𝛼𝛼𝛼𝛼(𝑘𝑘) ±𝛽𝛽𝛽𝛽(𝑘𝑘)

Theorem 2: If 𝑢𝑢(𝑥𝑥) =𝑥𝑥𝑚𝑚 𝑡𝑡ℎ𝑒𝑒𝑒𝑒 𝑈𝑈( 𝐾𝐾) = 𝛿𝛿(𝑘𝑘 − 𝑚𝑚) 𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝛿𝛿(𝑘𝑘 − 𝑚𝑚) =�1, 𝑖𝑖𝑖𝑖𝑘𝑘=𝑚𝑚

0, 𝑖𝑖𝑖𝑖𝑘𝑘 ≠ 𝑚𝑚

Theorem 3: If 𝑢𝑢(𝑥𝑥) = 𝑒𝑒𝑥𝑥 𝑡𝑡ℎ𝑒𝑒𝑒𝑒 𝑈𝑈(𝑘𝑘) = 1

𝑘𝑘!

Theorem 4: If u (x) = g (x) h (x) then 𝑈𝑈(𝑘𝑘) = ∑𝑘𝑘_𝑙𝑙₌₀𝛼𝛼(𝑙𝑙)𝛽𝛽(𝑘𝑘 − 𝑙𝑙)

Theorem 5: If 𝑦𝑦(𝑥𝑥) = 𝑦𝑦1(𝑥𝑥) 𝑦𝑦2(𝑥𝑥) 𝑡𝑡ℎ𝑒𝑒𝑒𝑒𝑌𝑌(𝑘𝑘) = ∑𝑘𝑘_𝑘𝑘₁₌₀𝑌𝑌₁(𝑘𝑘1) 𝑌𝑌₂(𝑘𝑘 − 𝑘𝑘1)

Theorem 6: If

(

)

1

(

)

,

n n

dx

x

y

d

x

y

=

then ₍ ₎

! )! ( )

( Y1 k n

k n k k

Y = + +

Theorem 7: If

y

(

x

)

=

e

λx then

λ

,

λ

!

)

(

k

Y

k

=

is constant

Theorem 8: If y (𝑥𝑥) = sin(𝑤𝑤𝑥𝑥+𝛼𝛼) 𝑡𝑡ℎ𝑒𝑒𝑒𝑒 𝑌𝑌 (𝑘𝑘) = 𝑤𝑤𝑘𝑘

𝑘𝑘!sin(

𝑘𝑘𝑘𝑘

2 +𝛼𝛼 ), where 𝛼𝛼,𝑤𝑤are constant.

Theorem 9: If (𝑥𝑥) = cos(𝑤𝑤𝑥𝑥+𝛼𝛼) 𝑡𝑡ℎ𝑒𝑒𝑒𝑒 𝑌𝑌 (𝑘𝑘) = 𝑤𝑤𝑘𝑘

𝑘𝑘!cos(

𝑘𝑘𝑘𝑘

2 +𝛼𝛼 ), where 𝛼𝛼,𝑤𝑤are constant.

ANALYSIS OF METHOD

Consider an nth order boundary value problem of the form

1

0 ),

,...,

,

(

)

(

x

=

f

x

y

′

y

′′

y

−1

≤

x

≤

y

n n (4) with the initial conditions













− − 1 1 2 2

,...

,

_n n

dx

y

d

dx

y

d

dx

dy

y

B

(5)

The differential transform of (4) is

(

_{) ( )}

!

)

(

n

k

F

n

k

Y

+

=

+

(6)

where

F

( )

k

is the differential transform of

f

(

x

,

y

,

y

′

,

y

′′

,...,

y

n−1

)

.

The transformed boundary condition can be written as

( )

,

(

)

(

)

(

)

,

(

)

0 1 1

n

m

B

k

Y

i

k

m

Y

A

k

Y

_m N k m i

<

=

−

=

∑∏

= − =

, (7)

where m is the order of the derivative in the boundary conditions and A,

B

_mare real constants.

Using (6) and (7), the value of

Y

(

i

),

i

=

1 ,

2 ,

3 ,...

can be determined and by using inverse differential transformation, the following approximate solution up

O

( )

x

N+1 can be determined

(3)

ANALYSIS OF METHOD FOR NONLINEAR HIGHER FRACTIONAL ORDER BOUNDARY VALUE PROBLEMS

Consider an nth order boundary value problem of the form

𝑦𝑦𝑒𝑒₍_𝑥𝑥_{) =}_𝑒𝑒−𝑥𝑥_𝑦𝑦−_𝑚𝑚1₍_𝑥𝑥₎

is given by

𝑌𝑌(𝑘𝑘+𝑒𝑒) =₍_𝑘𝑘₊𝑘𝑘!_𝑒𝑒_)!�� 1 �_{𝑚𝑚 −}1 1�𝑒𝑒� �

(−1)𝐾𝐾

𝐾𝐾! � 𝐾𝐾!𝑌𝑌(𝑘𝑘)�,𝑘𝑘= 0,2,4 …

𝑌𝑌(𝑘𝑘+𝑒𝑒) =₍_𝑘𝑘₊𝑘𝑘!_𝑒𝑒_)!�� 1

�1−_𝑚𝑚1�𝑒𝑒� �

(−1)𝐾𝐾

𝐾𝐾! � 𝐾𝐾!𝑌𝑌(𝑘𝑘)�,𝑘𝑘= 1,3,5, … (9)

Example 1: Consider a linear sixth order boundary value problem

6

( )

( ) 6 , 0

x

1,

y x

=

y x

−

e

< <

x

(10)

with the boundary conditions

(0)

1, (1)

0,

(0)

0,

(1)

,

(0)

1,

(1)

2 y

=

y

=

y

′

=

y

′

= −

e y

′′

= −

y

′′

= −

e

(11)

By applying theorems (6), (7) on the equation (10) we get,

6

1

6 (

6)

( )

!

(

)

n

Y k

k

n

=





+

=

_

−

_





+

∏

(12)

Differential Transformation of the boundary condition equation (11) is

0

1 (0)

1, (1)

0, (2)

,

( )

0

2

n

k

Y

Y k

=

= −

∑

=

0 0

( )

,

(

1) ( )

2

n n

k k

kY k

e

k k

Y k

e

= =

= −

−

= −

∑

(13)

Put

n

=

3 ,

k

=

0

in (6) we get

3

1 )

3 (

=

−

Y

, put

n

=

4 ,

k

=

0

in (6) we get

8

1 )

4 (

=

−

Y

Put

n

=

5 ,

k

=

0

in (6) we get

30

1 )

5 (

=

−

Y

, put

n

=

6 ,

k

=

0

in (6) we get

144

1 )

6 (

=

−

Y

Put

n

=

7 ,

k

=

0

in (6) we get

840

1 )

7 (

=

−

Y

, put

n

=

8 ,

k

=

0

in (6) we get

5760

1 )

(

k

=

−

Y

By using inverse differential transformation equation (3) and using boundary conditions we obtain the following series solution up to

O

(

x

8

)

,

5860

840

144

30

8

3

2

1 )

(

)

(

8 7

6 5 4 3 2 8

0

x

k

Y

x

y

k

=

−

=

∑

=

−

(4)

In the following table we compare the exact solution and DTM solution,

x

Exact solution DTM solution Error

0 1 1 0

0.1 0.994350367 0.996320492 1X10-3 0.2 0.976526077 0.976309706 2 X 10-4 0.3 0.944036589 0.944901165 8X10-5 0.4 0.894002981 0.895094824 1X10-3 0.5 0.823103881 0.822436068 1X10-3 0.6 0.727514353 0.728847758 1X10-3 0.7 0.602836803 0.604126777 1X10-3 0.8 0.444022296 0.445111434 1X10-3 0.9 0.245255774 0.245969802 1X10-4 1 0 0.000002480 1X10-5

Example 2: Consider an eighth order boundary value problem having fractional order with nonlinear function

8

( )

x

( )

y x

=

e

−

y x

₍₁₄₎

with the boundary conditions

(0)

1, (1)

2,

(0)

1,

(1)

,

(0)

1,

y

=

y

= −

y

′

=

y

′

=

e y

′′

=

1 (1)

2,

(0)

1,

(1)

6 y

′′

=

y

′′′

=

y

′′′

= −

₍₁₅₎

By applying theorems (6), (7) on the equation (14) and using (9) and the transformation of (15) becomes,

𝑖𝑖𝑖𝑖 𝑒𝑒= 1,𝑚𝑚=1₂

𝑌𝑌(𝑘𝑘+ 1) =₍_𝑘𝑘_{+ 1)!}𝑘𝑘! �� 1 �_{𝑚𝑚 −}1 1��

(−1)𝐾𝐾

𝐾𝐾! � 𝐾𝐾!𝑌𝑌(𝑘𝑘)�,𝑘𝑘= 0,2,4 …

𝑌𝑌(𝑘𝑘+ 1) =₍_𝑘𝑘_{+ 1)!}𝑘𝑘! �� 1 �1−_𝑚𝑚�1 � �

(−1)𝐾𝐾

𝐾𝐾! � 𝐾𝐾!𝑌𝑌(𝑘𝑘)�,𝑘𝑘= 1,3,5, … 𝑌𝑌(0) = 1

𝑌𝑌(𝑘𝑘+ 8) =₍_𝑘𝑘_{+ 8)!}𝑘𝑘! �� 1 �_{𝑚𝑚 −}1 1�8� �

(−1)𝐾𝐾

𝐾𝐾! � 𝐾𝐾!𝑌𝑌(𝑘𝑘)�,𝑘𝑘= 0,2,4 …

𝑌𝑌(𝑘𝑘+ 8) =₍_𝑘𝑘_{+ 1)!}𝑘𝑘! �� 1 �1−_𝑚𝑚�1 8� �

(−1)𝐾𝐾

𝐾𝐾! � 𝐾𝐾!𝑌𝑌(𝑘𝑘)�,𝑘𝑘= 1,3,5, …

0 0.2 0.4 0.6 0.8 1 1.2

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

(5)

3

6 9

13

0, (1)

2,

,

1, (2)

2,

2, (3)

1.333333333,

3, (4)

0.111111111,

4, (5)

1.851851852 10 ,

5, (6)

5.144032922 10 ,

6, (7)

2.041282905 10 ,

7, (8) 1.012541124 10

,

8, (9)

if

k

Y

if k

Y

f

k

Y

if

k

Y

if

k

Y

X

f

k

Y

X

if

k

Y

X

if

k

Y

X

if

k

Y

−

− −

−

=

= −

=

= −

=

= −

=

= −

=

= −

19

25 31

5.580583795 10

,

9, (10)

3.075718582 10

,

10, (11)

1.695171176 10

X

if

k

Y

X

if

k

Y

X

−

− −

=

= −

By using inverse differential transformation equation and using boundary conditions we obtain the following series solution up to

O

(

x

11

)

,

9

0

( )

k

y x

Y k x

=

−

=

∑

= −

1 2

x

+

2 x

2

−

1.333333333

x

3

+

0.111111111

x

4

−

1.851851852 10

X

−3

x

5

+

5.144032922 10

X

−6

x

6

−

2.041282905 10

X

−9

x

7

+

1.012541124 10

X

−1

x

38

−

5.580583795 10

X

−19

x

9

+

3.075718582 10

X

−25

x

10

−

1.695171176 10

X

−31

x

11

CONCLUSION

In this paper, the differential transformation method has been applied to obtain the numerical solution of linear and nonlinear higher order boundary value problems. The present method has been applied in a direct way without using linearization, discretization, or perturbation. By increasing the order of approximation more accuracy can be obtained. Comparison of the numerical results with the existing technique shows that the present method is more accurate.

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Applied Mathematics and Computation, 154 (2004) 299-311 .

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Applied Mathematics and Computation 165 (2005) 63-79.

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